Difference between revisions of "1986 AIME Problems/Problem 10"
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== Problem == | == Problem == | ||
− | In a parlor game, the magician asks one of the participants to think of a three digit number (abc) where a, b, and c represent digits in base 10 in the order indicated. The magician then asks this person to form the numbers < | + | In a parlor game, the magician asks one of the participants to think of a three digit number <math>(abc)</math> where <math>a</math>, <math>b</math>, and <math>c</math> represent digits in base <math>10</math> in the order indicated. The magician then asks this person to form the numbers <math>(acb)</math>, <math>(bca)</math>, <math>(bac)</math>, <math>(cab)</math>, and <math>(cba)</math>, to add these five numbers, and to reveal their sum, <math>N</math>. If told the value of <math>N</math>, the magician can identify the original number, <math>(abc)</math>. Play the role of the magician and determine <math>(abc)</math> if <math>N= 3194</math>. |
== Solution == | == Solution == | ||
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<cmath>m\equiv -3194\equiv -86\equiv 136\pmod{222}</cmath> | <cmath>m\equiv -3194\equiv -86\equiv 136\pmod{222}</cmath> | ||
− | This reduces <math>m</math> to one of 136, 358, 580, 802. But also <math>a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14</math> so <math>a+b+c\geq 15</math>. | + | This reduces <math>m</math> to one of <math>136, 358, 580, 802</math>. But also <math>a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14</math> so <math>a+b+c\geq 15</math>. |
Of the four options, only <math>m = \boxed{358}</math> satisfies this inequality. | Of the four options, only <math>m = \boxed{358}</math> satisfies this inequality. | ||
===Solution 2 === | ===Solution 2 === | ||
− | As in Solution 1, <math>3194 + m | + | As in Solution 1, <math>3194 + m \equiv 222(a+b+c) \pmod{222}</math>, and so as above we get <math>m \equiv 136 \pmod{222}</math>. |
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We can also take this equation modulo <math>9</math>; note that <math>m \equiv a+b+c \pmod{9}</math>, so | We can also take this equation modulo <math>9</math>; note that <math>m \equiv a+b+c \pmod{9}</math>, so | ||
<cmath>3194 + m \equiv 222m \implies 5m \equiv 8 \implies m \equiv 7 \pmod{9}.</cmath> | <cmath>3194 + m \equiv 222m \implies 5m \equiv 8 \implies m \equiv 7 \pmod{9}.</cmath> | ||
− | Therefore <math>m</math> is <math>7</math> mod <math>9</math> and <math>136</math> mod <math>222</math>. There is a shared factor in <math>3</math> in both, but Chinese | + | Therefore <math>m</math> is <math>7</math> mod <math>9</math> and <math>136</math> mod <math>222</math>. There is a shared factor in <math>3</math> in both, but the Chinese Remainder Theorem still tells us the value of <math>m</math> mod <math>666</math>, namely <math>m \equiv 358</math> mod <math>666</math>. We see that there are no other 3-digit integers that are <math>358</math> mod <math>666</math>, so <math>m = \boxed{358}</math>. |
== See also == | == See also == |
Revision as of 22:56, 2 April 2018
Problem
In a parlor game, the magician asks one of the participants to think of a three digit number where
,
, and
represent digits in base
in the order indicated. The magician then asks this person to form the numbers
,
,
,
, and
, to add these five numbers, and to reveal their sum,
. If told the value of
, the magician can identify the original number,
. Play the role of the magician and determine
if
.
Solution
Solution 1
Let be the number
. Observe that
so
This reduces to one of
. But also
so
.
Of the four options, only
satisfies this inequality.
Solution 2
As in Solution 1, , and so as above we get
.
We can also take this equation modulo
; note that
, so
Therefore is
mod
and
mod
. There is a shared factor in
in both, but the Chinese Remainder Theorem still tells us the value of
mod
, namely
mod
. We see that there are no other 3-digit integers that are
mod
, so
.
See also
1986 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.